Bounded derivative implies uniform continuity

Let $I\subseteq\mathbb{R}$ be an interval and let $f:I\to\mathbb{R}$ be differentiable on $I^\circ$.

Proposition: Suppose there exists $M\ge 0$ such that $|f'(x)|\le M$ for all $x\in I^\circ$. Then for all $x,y\in I$, $|f(x)-f(y)|\le M|x-y|.$ In particular, $f$ is Lipschitz on $I$ and hence uniformly continuous on $I$.

This proposition is one of the most common applications of the mean value theorem : derivatives control global oscillation.

Proof sketch: Fix $x<y$ in $I$ with $(x,y)\subseteq I^\circ$. By the mean value theorem, there exists $c\in(x,y)$ such that $f(y)-f(x)=f'(c)(y-x).$ Taking absolute values gives $|f(y)-f(x)|\le M|y-x|$. The same bound holds for $x>y$ by symmetry.